Math « Critacracy

On Collisions: Math and Colloquial Speech

November 5, 2009

So I was remembering this time in, say, fourth grade when the teacher posed the following question:

How many numbers are between ten and twenty?

Of course, this problem is well within the skills of a fourth-grader. Simply subtract ten from twenty and you get ten. In fact, the only appreciable difficulty is in the interpretation of the question. Students, particularly children, struggle with these so-called “word-problems”: problems where the math is relatively simple, but the problem is phrased as a question. You know, using words and stuff.

Being who I am, I completely failed at understanding the question; I answered, “Eleven.”

So I got laughed at… But to this day I maintain that 11 is just as accurate as—indeed more accurate than—10. Why? It all lies in the interpretation of the word “between.” If “between 10 and 20″ means

$\{x\in\mathbb{N}\, |\, 10 < x \le 20\}$

Then yes, there are 10 such numbers: $11, 12, 13, 14, 15, 16, 17, 18, 19, 20$ . But who the hell uses “between” to mean “including the upper limit but excluding the lower limit”? In my opinion, there are only two reasonable ways to interpret “between ten and twenty”:

$\{x\in\mathbb{N}\, |\, 10 \le x \le 20\}$
$\{x\in\mathbb{N}\, |\, 10 < x < 20\}$

But then the correct answer is either $9$ or $11$ . I chose option (1) in fourth grade, but option (2) is perfectly reasonable.

The problem is that this becomes no longer strictly a subtraction problem. What the teacher wants is for the students to compute $20 - 10 = 10$ . But this is wrong, so how do we reinterpret the question?

I propose the following: first consider the question, “what are the numbers between ten and twenty?” This question still has the ambiguity of the word “between,” but no person in his right mind would answer:

$10, 11, 12, 13, 14, 15, 16, 17, 18, 19$

$11, 12, 13, 14, 15, 16, 17, 18, 19, 20$

Then we can ask, after the set of numbers has been identified, “how many of them are there?” I think if the problem were posed this way, no one would ever arrive at an answer of eleven (unless they miscount).

Further Rambling

I’ve decided that word problems like the ones we’re given in grade school don’t really count as “math.” I think they’re physics. Consider the topics covered in college- and higher-level mathematics. They’re abstract and almost completely disconnected from the real world. A mathematician doesn’t care if we live in Euclidean space or a Minkowski space; if they can prove something about both, then that is interesting.

A physicist (or statistician maybe) concerns himself with questions like these word problems—problems dealing with possible real-world scenarios. When we answer such questions, we don’t really learn a lot about math; rather, we learn about the connection between math and the world we live in.

That’s not to say word problems aren’t good or helpful, but I think school teachers should be very careful both in posing questions and in receiving answers. What if my teacher had let me explain my logic? It would have been nice to shut up the kids who laughed at me for a “stupid” answer, and they might learn something at the same time.

Leave a Comment » | Math, Social Commentary | Tagged: Between, Pedantry, Semantics | Permalink
Posted by Jay

A Logical Fallacy

August 17, 2009

Most people are familiar with the “gambler’s paradox.” You know, the statistical paradox regarding, for example, the flip of a coin: you flip a coin 99 times, and each time it comes up heads. On the 100th flip, what is the probability that the coin shows tails?

It’s called a paradox because intuitively you want tails to be more likely than heads on the 100th flip, but this obviously isn’t the case; the probability of tails is 50% no matter what has happened in the past.

But despite people’s willingness to accept the mathematics behind the trivial example above, many still fall for the gambler’s paradox in more complicated situations. For example, I was talking to a friend the other day who said something like this:

“I was on an airplane to Florida and I swear I saw a plane crash. The only thing that prevented me from being totally scared was the fact that because they crashed, I must be less likely to be in a crash.”

I responded with, “Wait… What?” I tried to explain that her logic made no sense; if one plane has crashed, another plane isn’t suddenly less likely to crash just because it needs to be in keeping with a statistical average.

Suppose every day 1 in 10 planes crashes (obviously not true, but I’m too lazy to look up the real figures). Then the chances of two planes crashing in one day is .10 x .10 = .01, or 1%. Here is the source of the fallacy. The probability of two planes crashing is lower than the probability of one plane crashing. However, the fact that one plane has crashed has no bearing on the probability of another plane crashing.

That is, if one plane has crashed, the probability of it crashing is 100%. Therefore the probability of that plane crashing and your plane crashing is 1 x .10 = .10, or 10%. This is, of course, if we assume that the two probabilities are independent.

In reality, the probabilities are somewhat dependent. If one plane crashes, pilots on other planes might be informed and would then change the way they fly, which, in turn, would change the probability of a subsequent crash. Nevertheless, the impact is probably small, and we can safely assume independence.

Leave a Comment » | Math | Tagged: Gambler's Paradox, Math, Plane Crashes, Probability, Statistics | Permalink
Posted by Jay

The Monty Hall Problem

July 17, 2009

In a rare coincidence, I spent a solid two hours considering the Monty Hall problem the other night, and today I was linked to a rather absurd discussion on the very same topic. In this post I plan to present not a new solution to the problem, but rather a new explanation that makes the most intuitive sense to me.

Here’s a quick review: the Monty Hall problem is a probability problem based on a game show. There are three closed doors. Behind one is a prize (like a car), and behind the other two are booby prizes (like goats or pigeons or something). The host asks you to choose a door—A, B, or C. After you’ve chosen a door, the host opens a different door and reveals a booby prize. You are then asked whether you want to change your mind or not. What should you do?

The answer, it would appear, is that it doesn’t matter whether you change your mind or not. However, after careful consideration, you can see that you’re better off changing your mind (!)

What does that mean? It means that if you choose door B, and the host opens door A, then you should choose door C. The discussion I linked to above and the Wikipedia provide some useful and interesting explanations. What follows is the explanation I thought up.

My Explanation

My thought process stems from one question: how is it that “changing your mind” can affect the situation? Consider what happens if you choose door A and the host opens door B. There are exactly TWO ways to arrive at this situation:

You have chosen incorrectly (the car is behind door C)
You have chosen correctly (the car is behind door A)

A common (and correct) explanation is that scenario 1 is indeed more likely than scenario 2. But also consider this: if the car is behind door A and you choose door A, then the host can open either door B or door C. If you did a statistical analysis—for example, have 300 contestants choose door A—you would find, on average, ~33% would have chosen correctly (because the car has equal probability of being behind A, B, and C), and ~66% would have chosen incorrectly. Suppose the correct door is door A.

Those ~33% who chose correctly, only half of them, or ~16.5% of the who group, would see door B open as in the situation I described. The other 16.5% would see door C open. For the 33% who chose door C, ALL of them will see door B open (because door A is correct). The 33% who chose door B would NOT see door B open, and would those find themselves in a different situation altogether.

Adding, we find that 16.5% + 33% = 49.5% (actually it’s 50% without rounding) of the entire sample will see door B open if door A is correct–as expected. However, of those 50%, 33% chose incorrectly, and only 16.5% chose correctly. Therefore, 2/3 of the time, it’s better to switch.

You can now easily take this explanation and let door B or door C be correct, and you will find the same result.

The Meta-Explanation

What I like about my explanation is that it can answer the question I first posed: how can changing your mind affect the situation in such a way? The answer: it can’t. My explanation describes how there are only so many ways to arrive at a particular “situation,” and each situation is reached more frequently by people who have chosen incorrectly.

If you examine my argument closely, you’ll find it isn’t much more than an expansion of the standard “you’re more likely to choose wrong” argument, but I like it because it helps me understand the solution more intuitively.

Leave a Comment » | Math | Tagged: Math, Monty Hall, Probability | Permalink
Posted by Jay

WolframAlpha: Second Impressions

May 31, 2009

After I took out some of my aggression on WolframAlpha a couple weeks ago (and managed to misspell “Stephen Wolfram” and not even notice until now), I’ve decided to somewhat revise my statements.

While WolframAlpha is still nothing special, I’ve been using it pretty much every day. It’s sort of a replacement for my calculator, much in the way Google was. I used to do, for example, physics problem sets and type basic equations into Google, like $1.5 m * sin(57^\circ) \, in \, feet$ . WolframAlpha has filled this role because it can do things like take $x^2 - 3x +1$ and find roots.

It can even somtimes surprise you with its abilities:

Not bad

So the moral of the story is that WolframAlpha isn’t all bad; we sciency-types can still find it somewhat useful.

In other news, wasn’t Google Squared supposed to be out by now?

1 Comment | Math, Technology | Tagged: Matrices, WolframAlpha | Permalink
Posted by Jay

The Factor Theorem

April 28, 2009

I was a little angry the other day to be assigned the following problem for homework:

Suppose $a_o,\dots,a_n$ are in $\mathbb{C}.$ Show that $\displaystyle p(z) = a_nz^n+a_{n-1}z^{n-1}+\cdots+a_0 = \prod_{i=1}^n (z-z_i)$ for $z_i$ (not necessarily distinct) in $\mathbb{C}.$ Assume the fundamental theorem of algebra.

Why the anger? Because I don’t know algebra; I’m not taking algebra. I found proofs on the internet, but they all require algebra that I simply don’t know. So I resorted to the following:

Lemma

If $p(z)$ is a polynomial of degree $n$ with $p\!:\!\mathbb{C}\rightarrow \mathbb{C},$ then the Taylor polynomial of $p$ at $a$ is equal to $p(z).$

The proof of this statement is obvious and can be found in a couple of ways: show that the two polynomials are equal up to order $n$ and therefore equal; find the $k^{th}$ derivative of $p$ at $a$ and use that to show that $a_k$ is also the coefficient for the $k^{th}$ term of the Taylor polynomial.

Theorem

Suppose $a_o,\dots,a_n$ are in $\mathbb{C}.$ Then $\displaystyle p(z) = a_nz^n+a_{n-1}z^{n-1}+\cdots+a_0 = \prod_{i=1}^n (z-z_i)$ for $z_i$ (not necessarily distinct) in $\mathbb{C}.$

Proof

Given $p(z) = a_nz^n+a_{n-1}z^{n-1}+\cdots+a_0$ then

$\displaystyle p(z) = \sum_{k=0}^n\! \frac{p^{(k)}(a)}{k!}(z-a)^k$ by the lemma. That is,

$\displaystyle p(z) = p(a) + \sum_{k=1}^n\! \frac{p^{(k)}(a)}{k!}(z-a)^k$

$\displaystyle p(z) = p(a) + (z-a) \sum_{k=1}^n\! \frac{p^{(k)}(a)}{k!}(z-a)^{k-1}$

Suppose $p(a) = 0$ . That is, suppose $a = z_1$ where $z_i$ are the roots of $p(z)$ . We know $z_1$ must exist by the fundamental theorem of algebra.

$p(z) = 0 + (z-a) \displaystyle\sum_{k=1}^n\! \frac{p^{(k)}(a)}{k!}(x-a)^{k-1} = q(z)$

Clearly, $\displaystyle \frac{p(z)}{z-a} = q(z)$

We continue in the same way; take a root $z_2$ of $q(z)$ and repeat the above by taking the Taylor polynomial of $q$ at $a = z_2$ . This will eventually reduce the equations to:

$\displaystyle \frac{p(z)}{(z-z_1)(z-z_2)\cdots(z-z_n)} = \frac{q(z)}{(z-z_2)(z-z_3)\cdots(z-z_n)}$

$\displaystyle = \cdots = \frac{z-z_n}{z-z_n} = 1$

Considering only the first and last terms, it is clear that

$\displaystyle p(z) = (z-z_1)(z-z_2)\cdots(z-z_n) = \prod_{i=1}^n (z-z_i)$

QED

Leave a Comment » | Math | Tagged: Calculus, Factor Theorem, Fundamental Theorem of Algebra, Math | Permalink
Posted by Jay

Who Has It Better?

February 17, 2009

We were shown the below picture in one day in physics. It’s a 3D-rendering of the magnetic field due to a solenoid. It was heroically calculated by a computer using the Biot-Savart law. I think it’s fantastic.

But I started thinking about all the cool pictures that mathematicians make. There are some obvious ones, like Budhhabrot and other colored fractals. And there are even simpler ones, like domain-colored graphs of functions $f\!:\mathbb{C}\!\rightarrow\!\mathbb{C}$

Leave a Comment » | Math | Tagged: Biot-Savart, Buddhabrot, Domain Coloring, Fractals, Magnetic Fields, Solenoids | Permalink
Posted by Jay

Roulette Math – Part II

October 7, 2008

Thanks to no one, I’ve finally reached a conclusion to the question I posed at the end of my Roulette Math post. To prove it, proof by induction with two variables will be used. Explicitly,

let $P(n,k)$ be the statement $n^k = 1 + (n-1)\displaystyle\sum_{i=0}^{k-1} n^i$

To prove $P$ ,

Prove $P(1,1)$
Assume $P(n,1)$
Prove that $P(n+1,1)$ is true if $P(n,1)$ is true
Assume $P(n,k)$
Prove that $P(n,k+1)$ is true if $P(n,k)$ is true

Phase 1

$1^1 = 1 + (1-1)\displaystyle\sum_{i=0}^{1-1} n^i = 1 + 0 = 1$ is true

Phases 2, 3

Induction is not necessary here.

$1 + ((n+1)-1)\displaystyle\sum_{i=0}^{0} n^i = 1 + n = (n+1)^1$

Phases 4,5

Assume $n^k = 1 + (n-1)\displaystyle\sum_{i=0}^{k-1} n^i$ (phase 4)

All that remains is to show that $P(n,k+1)$ is true if the above is true (phase 5):
Since $n^k = 1 + (n-1)\displaystyle\sum_{i=0}^{k-1} n^i$ by assumption, then $n^k + (n-1)n^k = (n-1)n^k + 1 + (n-1)\displaystyle\sum_{i=0}^{k-1} n^i$

$n^k + (n-1)n^k = n^k(1+(n-1)) = n^k(n) = n^{k+1}$

$= (n-1)n^k + 1 + (n-1)\displaystyle\sum_{i=0}^{k-1} n^i$

So $n^{k+1} = (n-1)n^k + 1 + (n-1)\displaystyle\sum_{i=0}^{k-1} n^i$

$n^{k+1} = 1 + (n-1)n^k + (n-1) + (n-1)n +...+ (n-1)n^{k-1}$

$= 1 + (n-1) + (n-1)n +...+ (n-1)n^{k-1} + (n-1)n^k$

$= 1 + (n-1)\displaystyle\sum_{i=0}^{k} n^i$

Thus, if $P(n,k)$ is true, $P(n,k+1)$ must be true. Phase 5 complete. QED.

Further Application To Roulette

Once again, imagine you’re playing roulette, betting only on red, black, even, or odd (the ones with 2:1 payout). Instead of doubling, you decide that you’d rather triple, quadruple, or even $n$ -uple your bet each round. You continue multiplying your principle bet, $p$ by $n$ until you win. After $k$ rounds, you have lost

$p+p(n)+p(n^2)+...+p(n^k) = p\displaystyle\sum_{i=0}^k n^i$ dollars

If you win in round $k+1$ you will win $p(n^k) = p + p(n-1)\displaystyle\sum_{i=0}^{k} n^i$

Your total earnings are what you’ve won minus what you spent:

$p + p(n-1)\displaystyle\sum_{i=0}^{k} n^i - p\displaystyle\sum_{i=0}^k n^i$

Which, according to Mathematica, can be simplified to

$\frac{p(1+n^{k+2}-2n^{k+1})}{n-1}$

That means if your initial bet is $100, you quintuple your bet every time, and it takes you 4 rounds to win, you’ll walk away with a bonus $103,700. That is, if you can afford the betting (and if there’s no maximum bet!)

Leave a Comment » | Math | Tagged: Math, Roulette | Permalink
Posted by Jay

Roulette Math

July 29, 2008

This problem and subsequent math came up after a discussion I had with a friend about roulette. Couldn’t you, he wondered, just always bet on red and keep doubling your bet until you win? Aren’t you guaranteed to make money that way?

Short answer: yes. You will earn exactly the amount of money you placed for your initial bet. I though through the math of this, and arrived at this conclusion after a couple hours of pondering and scribbling. Why exactly does this method work?

It’s true because the following is true:

$2^k = 1 + \displaystyle\sum_{i=0}^{k-1} 2^i$

I will return to roulette in a moment, but first let me offer a proof for the above statement:

Proof: This is a proof by induction. We begin by checking to see if the equation holds for $k = 1$ and we see that it does:

$2^1 = 1 + \displaystyle\sum_{i=0}^0 2^i$

$2 = 1 + 1 = 2$

We continue by assuming the equation holds for any arbitrary $k$ . We then must show that $k \Rightarrow k+1$ :

$2^{k+1} = 1 + \displaystyle\sum_{i=0}^{k} 2^i$

We do this by adding $2^k$ to both sides.

$2^k + 2^k = 2^{k+1} = 1 + \displaystyle\sum_{i=0}^{k-1} 2^i + 2^k$

$2^{k+1} = 1 + 2^0 + 2^1 + ... + 2^{k-1} + 2^k$

$2^{k+1} = 1 + \displaystyle\sum_{i=0}^{k} 2^i$

QED

Application To Roulette

I’m dying, btw, to have an actual mathematician check my math and tell me if the above proof is valid. But on to the money! The above theorem essentially says, in english, that any power of two is the sum of the previous powers, plus one. We apply that to gambling thusly:

Imagine you play roulette, betting only on red (or black/even/odd), and begin with a principal amount of cash, $p$ . After $k$ rounds of doubling your bet, betting, and losing, you have lost

$p + 2p + p(2^2) + ... + p(2^k)$ or:

$p\displaystyle\sum_{i=0}^k 2^i$

If you win on the next round, round $k+1$ , you will win of course $p(2^{k+1})$ . From the theorem, it follows that you have won

$p(2^{k+1}) = p + p\displaystyle\sum_{i=0}^k 2^i$

If we subtract what we’ve won from what we’ve spent, we find how much we’ve gained:

$p + p\displaystyle\sum_{i=0}^k 2^i - p\displaystyle\sum_{i=0}^k 2^i = p$

We find that we are left having gained exactly $p$ . Incidentally, I’ve discovered that the theorem I proved above can (probably) be extended to all bases:

$n^k = 1 + (n-1)\displaystyle\sum_{i=0}^{k-1} n^i$

But I haven’t really found a way of proving it. I’m quite certain that it can be proved using similar logic as before. That is, you add $(n-1)$ -many $n^k$ ’s to each side. I just have no way of formalizing it. Any help?

2 Comments | Math | Permalink
Posted by Jay

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